Calculus 3
Lia Vas
Parametric Surfaces. Substitution
Recall that a curve in space is given by parametric equations as a function of single parameter t
x = x(t)
y = y(t)
z = z(t).
A curve is a one-dimensional object in space so its parametrization is a function of one variable.
Analogously, a surface is a two-dimensional object in space and, as such can be described using
two variables. If a surface is given by an explicit equation z = f (x, y) the dependence on two
independent variables is clearly visible. However, a parametric representation of an implicit surface
F (x, y, z) = 0 may often be very useful. In particular, it may be useful to parametrize such surface
using two parameters possibly different from x and y. In particular, a surface given by the parametric
equations
x = x(u, v) y = y(u, v) z = z(u, v)
is referred to as a parametric surface and the two independent variable u and v as parameters.
The equations above are called the parametric equations of the surface. Thus, we have the
following.
Curve
Surface
Dimension Parameter(s) Equations
1
t
~r(t) = hx(t), y(t), z(t)i
2
u, v
~r(u, v) = hx(u, v), y(u, v), z(u, v)i
Examples.
√
A surface of revolution z = f ( x2 + y 2 )
can be parametrized using polar coordinates as
follows
x = r cos θ,
y = r sin θ,
z = f (r).
In particular, the following surfaces of revolution
are parametrized in this way.
1. The paraboloid z = x2 + y 2 has parametric
representation by x = r cos θ, y = r sin θ,
z = r2 .
√
2. The cone z = x2 + y 2 has parametric representation by x = r cos θ, y = r sin θ, z = r.
3. The upper hemisphere
of the sphere x2 +y 2 +z 2 = 9 has parametric representation by x = r cos θ,
√
y = r sin θ, z = 3 − r2 .
A cylindrical surface obtained from a curve in one of the coordinate planes can be parametrized
using the curve parametrization and the remaining variable as the second parameter. For example,
consider the cylinders below.
4. The cylinder x2 + y 2 = 4 is based on the circle x = 2 cos t, and y = 2 sin t. The value of
the z-coordinate of a point on the cylinder
represents the height h of that point. So,
the cylinder can be parametrized as
x = 2 cos t, y = 2 sin t,
z=h
using the angle t and the height h as two
parameters.
5. Similarly, the cylinder y 2 + z 2 = 4 is based on the circle y = 2 cos t, and z = 2 sin t in yz-plane.
The value of the x-coordinate of a point on the cylinder can also be considered to be the height.
Using h for that value, the cylinder can be parametrized as
x = h,
y = 2 cos t, z = 2 sin t.
Tangent Plane and Surface Area
The derivative vectors hxu , yu , zu i and
hxv , yv , zv i are in the tangent plane of a parametric surface x = x(u, v), y = y(u, v),
z = z(u, v), (also written shortly as ~r =
hx(u, v), y(u, v), z(u, v)i). Thus, the tangent
plane of the surface has the perpendicular vector
→
−
−
ru × →
rv = hxu , yu , zu i × hxv , yv , zv i.
In case when the surface z = z(x, y) is given
by the parameters u = x and v = y, this reduces
to the familiar formula (−zx , −zy , 1).
The surface area S of this surface over the region D in uv-plane can be obtained by integrating
surface area elements dS over sub-rectangles of region D. The area of each element dS can be approximated with the area of the parallelogram in the tangent plane just as before when we considered
surfaces parametrized using x and y.
Recall that the area of a parallelogram formed by two vectors is the length of their cross product.
Thus,
−
−
dS = |hxu , yu , zu i × hxv , yv , zv i|dudv or, written shortly, dS = |→
ru × →
rv |dudv.
The formula for the total surface area S can be obtained by integrating dS over entire region D.
Thus
Z Z
Z Z
Z Z
→
−
→
−
S=
dS =
| ru × rv | dudv =
|hxu , yu , zu i × hxv , yv , zv i| dudv.
D
D
D
In the case when the surface is parametrized by u = x and v = y as z = z(x, y) this formula
becomes the one we have encountered before.
S=
Z Z q
D
1 + zx2 + zy2 dxdy
General substitution for double integrals.
In many cases, a region over which a double integral is being taken may have easier representation
in another coordinate system, say in uv-plane, than in xy-plane. In cases like that, one can transform
the region in xy-plane to a region in uv-plane by the substitution
x = g(u, v) y = h(u, v).
Thus, a substitution is just a convenient reparametrization of a surfaceR zR = f (x, y) when the
parameters x and y are changed to u and v. When evaluating the integral
D f (x, y)dxdy using
substitution, the area element dA = dxdy becomes |J|dudv where the Jacobian determinant J is
given by
x
∂x ∂x u xv ∂u ∂v .
J = ∂y ∂y = yu yv ∂u
∂v
Thus,
Z Z
f (x, y) dx dy =
D
Z Z
f (x(u, v), y(u, v)) |J|du dv
D
Note that in one-dimensional case, the Jacobian determinant is simply the derivative of the
substitution u = u(x) solved for x so that x = x(u) ⇒ dx = x0 (u)du.
Jacobian for polar coordinates. The polar coordinates x = r cos θ and y = r sin θ can
be considered as a substitution in which u = r and v = θ. Thus, xr = cos θ, xθ = −r sin θ and
yr = sin θ, yθ = r cos θ. The Jacobian is
xr xθ cos θ −r sin θ
J=
=
yr yθ sin θ r cos θ
= r cos2 θ + r sin2 θ = r.
This explains the presence of r in the integrals of the section on Polar Coordinates.
Z Z
f (x, y)dxdy =
Z Z
f (r cos θ, r sin θ) r dr dθ.
D
Practice problems.
1. Find the equation of the tangent plane to given parametric surface at the specified point.
a) x = u + v
y = 3u2
z = u − v;
(2, 3, 0)
b) x = uv
y = uev
z = veu ;
(0, 0, 0)
2. Find the area of the surface.
a) The part of the plane z = x + 2y that lies above the triangle with vertices (0,0), (1,1) and
(0,1).
b) The part of the surface z = y 2 + x2 that lies between the cylinders x2 + y 2 = 1 and
x2 + y 2 = 4. Write down the parametric equations of the paraboloid and use them to find
the surface area.
√
c) The part of the cone z = x2 + y 2 that lies between the cylinders x2 + y 2 = 4 and
x2 + y 2 = 9. Write down the parametric equations of the cone first. Then find the surface
area using the parametric equations.
3. Consider the cylinder x2 + z 2 = 4.
a) Write down the parametric equations of this cylinder.
b) Using the parametric equations, find the tangent plane to the cylinder at the point (0, 3, 2).
c) Using the parametric equations and formula for the surface area for parametric curves,
show that the surface area of the cylinder x2 + z 2 = 4 for 0 ≤ y ≤ 5 is 20π.
4. Use the given substitution to evaluate the integral.
dx dy where D is the region bounded by the lines y = x, y = x − 2, y = −2x,
and y = 3 − 2x. The substitution x = 13 (u + v), y = 31 (v − 2u) transforms the region to a
rectangle 0 ≤ u ≤ 2 and 0 ≤ v ≤ 3.
a)
RR
b)
RR
xy dx dy where D is the region in the first quadrant bounded by the curves y = x,
y = 3x, y = x1 , and y = x3 . The substitution x = uv , y = v transforms the region into a
√
√
region with bounds 1 ≤ u ≤ 3 and u ≤ v ≤ 3u.
c)
RR
D (3x + 4y)
D
xy dx dy where D is the region in the first q
quadrant bounded by the curves y = x,
√
1
3
y = 3x,y = x , and y = x . The substitution x = uv , y = uv transforms the region into
a square 1 ≤ u ≤ 3 and 1 ≤ v ≤ 3.
D
Solutions.
1. a) hxu , yu , zu i = h1, 6u, 1i and hxv , yv , zv i = h1, 0, −1i. The cross product of these vectors is
h−6u, 2, −6ui. Now we need to find u-value (v-value is not needed since v does not appear in
this product) that corresponds to the point (2, 3, 0). When (x, y, z) = (2, 3, 0), x = u + v = 2
and z = u − v = 0. From this we have that u = v = 1. So, the perpendicular vector is
h−6, 2, −6i. The equation of the plane is −6(x − 2) + 2(y − 3) − 6(z − 0) = 0 ⇒ 3x − y + 3z = 3
b) hxu , yu , zu i = hv, ev , veu i and hxv , yv , zv i = hu, uev , eu i. The cross product of these vectors is
heuv (1 − uv), veu (u − 1), uev (v − 1)i. Now we need to find the u and v values that correspond
to the point (0, 0, 0). When (x, y, z) = (0, 0, 0), uev = 0 ⇒ u = 0 and veu = 0 ⇒ v = 0. From
this we have that u = v = 0. So, the perpendicular vector is h1, 0, 0i. The equation of the plane
is 1(x − 0) + 0(y − 0) + 0(z − 0) = 0 ⇒ x = 0 (yz-plane).
RR q
2. a) You can use the standard xy-parametrization here and the formula S = T 1 + zx2 + zy2 dxdy
where T is the given triangle. z = x + 2y ⇒ zx = 1 and zy = 2. The bounds for x
are 0 and 1 and y is bounded by the line connecting (0,0) and (1,1) from below and by
the line that connects (0,1) and (1,1) from above. Thus x√ ≤ y ≤ 1. The surface area is
√ R
√
R R √
S = 01 x1 1 + 1 + 4 dxdy = 6 01 (1 − x)dx = 6(1 − 21 ) = 26 = 1.22.
b) The paraboloid can be parametrized by x = r cos t, y = r sin t, z = x2 + y 2 = r2 . hxr , yr , zr i =
hcos t, sin t, 2ri and hxt , yt , zt i = h−r sin t, r cos t, 0i.√The cross product of these √vectors is
h−2r2 cos t, −2r2 sin t, ri. The length of this product is 4r4 cos2 t + 4r4 sin2 t + r2 = 4r4 + r2 =
q
√
r2 (4r2 + 1) = r 4r2 + 1.
The bounds for the integration are determined by the projection in the xy-plane which is the
region between the circles
x2 +R y 2√= 1 and x2 + y 2 = 4. Thus 0 ≤ t ≤ 2π and 1 ≤ r ≤ 2. So,
R 2π
the surface area is S = 0 dt 12 r 4r2 + 1dr = 2π4.91 = 30.85.
√
c) The cone can be parametrized by x = r cos t, y = r sin t, z = x2 + y 2 = r. hxr , yr , zr i =
hcos t, sin t, 1i and hxt , yt , zt i = h−r sin t, r cos
cos t,
√ t, 0i. The cross2product of√these vectors
√ is h−r√
2
2
2
2
2
2
2
−r sin t, ri. The length of this product is r cos t + r sin t + r = r + r = 2r = 2r.
The bounds for the integration are determined by the projection in the xy-plane which is the
2
2
region between the circles
x2 +R y√
= 4 and x√
+ y 2 = 9. Thus
√ 0 ≤ t ≤ 2π and 2 ≤ r ≤ 3. So,
R 2π
the surface area is S = 0 dt 23 2rdr = 2π 2( 29 − 42 ) = 5π 2.
3. a) The cylinder can be parametrized as x = 2 cos t, y = y, z = 2 sin t.
b) hxt , yt , zt i = h−2 sin t, 0, 2 cos ti and hxy , yy , zy i = h0, 1, 0i. The cross product is h−2 cos t,
0, 2 sin ti. The t-value that corresponds to (0, 3, 2) can be obtained from x = 2 cos t = 0 and
z = 2 sin t = 2. Thus t = π2 . The perpendicular vector is h0, 0, 2i and the equation of the plane
is 0(x − 0) + 0(y − 3) + 2(z − 2) = 0 ⇒ z = 2.
√
c)
We
already
computed
the
cross
product
h−2
cos
t,
0,
2
sin
ti.
Its
length
is
4 cos2 t + 4 sin2 t =
√
4 = 2. TheR t-bounds
are 0 and 2π and the y-bounds are given to be 0 ≤ y ≤ 5. So, the surface
2π R 5
area is S = 0 0 2dydt = 2π(5)(2) = 20π.
1 1
RR
xu xv 3
3 = 1 + 2 = 1 .
= −2
4. a) Calculate the Jacobian J =
1
D (3x + 4y) dx dy =
9
9
3
yu yv 3
3
R2R3
4
1
1 R2
v2
4v 2
8uv 3
1 R2
9
0 0 (u + v + 3 (v − 2u)) 3 du dv = 3 0 (uv + 2 + 6 − 3 )|0 du = 3 0 (3u + 2 + 6 − 8u) du =
1
(6 + 9 + 12 − 16) = 11
3
3
1 −u RR
R 3 R √3u u 1
xu xv 2 √
b) The Jacobian is J =
= v v = v1 .
xy
dx
dy
=
D
1
u v v v du dv =
0 1 yu yv √
√
√ u2 3
√
R3
R3
√3u
3 2 |1 = 4 ln 3 = 2 ln 3 = 2.197
1 udu ln v| u = 1 udu ln 3 = ln
c) The Jacobian is J
R 3 R 3 qu
1
1
v
√
uv
1
2v
x
= u
yu
R3R3 u
du dv =
1
1 2v
xv
yv
=
du dv =
√ −√ u 2√ v 3 = 1 + 1
u
4v
4v
√
2 v
u2 3
| ln v|31 = 2 ln 3 = 2.197
4 1
√1
2 uv
√
√v
2 u
=
1
.
2v
RR
D
xy dx dy =